A Mathematical Emergency.

CraigD · 10-16-2008

Quote:

Originally Posted by Qfwfq

Actually I found some reasonably readable stuff, although I still don't agree with it.

Looking at some of the other pages there, I see what you mean. ‘On “Cohesive” Terms vs. “Arabic” Terms’, for example, has superscripts, subscripts, arrows, and boxes, but no fractions, and even uses $\log_T N$ rather than $\frac{\ln(N)}{\ln(T)}$ . It appears to be the scanned image of a typed and hand drawn paper document.

Don, ‘On “Cohesive” Terms vs. “Arabic” Terms’ is IMHO more readable than your several other non-handwritten papers. I disagree with its claim, however. Since it addresses your idea of “cohesive terms”, you might want to consider discussing it, rather than your various Beal’s & Fermat’s last conjecture proofs.

Don Blazys · 10-17-2008

To: CraigD.

Here's the difference between a "division by zero", and an "indeterminate form":

6/3=2 is valid because: 2*3=6. However,

6/0= N is "undefined" because: N*0 does not equal 6, and

0/0= N is "indeterminate" because: any number N*0=0.

Thus, if 0/0= N, where N can be "any number", then the expressions in my proof clearly work out to:

(1)^(0/0)=1, therefore, c(1)^(0/0)= c, and (c(1)^(0/0))^2=c^2.

In other words, we need not know the value of the "indeterminate form" (0/0) in order to determine that
(1)^(0/0)=1 because 1 raised to any power, (including any "indeterminate power" such as 0/0), still equals 1.

Moreover, the all important doctrine of maintaining logical/mathematical consistency requires that our results have the exact same meaning regardless of whether we first let z=1 and z=2, then "cross out" the logarithms or whether we first let T=c, then let z=1 and z=2 so as to beget those absolutely benign "indeterminate forms".

In short, the "indeterminate forms" in no way derail my proof.

From my humble point of view, the proof works perfectly, and I expect that the editors and referees at the Journal Of The American Mathematical Society will see what I, along with many others, see. They have certainly had it long enough! (My website includes a few of the many letters of recomendation that I recieved, including one from a famous NASA/JPL planetary scientist that is directed to the AMS.)

"Cohesive terms" is just a name that I gave my discovery. Thus, at this time, you will not find that name or the construct to which it pertains anywhere but in my writings. (My innate humility prevented me from calling them "Blazys terms".) In my opinion, they are the most beautifull, elegant, and powerfull construct in all of mathematics. So much so that I no longer view the "Beal Conjecture" and "Fermat's Last Theorem" as "famous conjectures", but as "conjectures that would never have been made, had mankind learned to represent algebraic terms correctly to begin with".

I am very proud of both you, and Qfwfq. Despite my not being able to write out my mathematical expressions and equations in a format that you are used to, you fought your way through and have now come to a point where you nearly understand my proof! I am also very gratefull for all your help and advice.

Don.

Qfwfq · 10-20-2008

Quote:

Originally Posted by Don Blazys

0/0= N is "indeterminate" because: any number N*0=0.

Thus, if 0/0= N, where N can be "any number", then the expressions in my proof clearly work out to:

(1)^(0/0)=1, therefore, c(1)^(0/0)= c, and (c(1)^(0/0))^2=c^2.

In other words, we need not know the value of the "indeterminate form" (0/0) in order to determine that
(1)^(0/0)=1 because 1 raised to any power, (including any "indeterminate power" such as 0/0), still equals 1.

Note Don that 0/0 is just as "undefined" as 6/0 is. The step from saying "any number N*0=0" to saying "0/0= N, where N can be 'any number'" is a paralogism; you should notice that the nexus involves dividing or multiplying by 0, so you draw your conclusions by an "as if" and quite circular argument.

The third assert is interesting because, while fraught with the formal woe of being an expression in which one term does not define a value, you justify the claim by the seemingly compelling argument that the whole expression is totally independent on the undefined term anyway, so it doesn't matter. Now, apart from the above difference between "can be any value" and "defines no value", even when loosely put in the former terms there remains the fact that "any value" can be $\infty$ too. Now this is a very subtle matter because the indepence can, at first glance, seem so intrinsic that one may be convinced by the argument. However one may use the rules of logarithms to write:

$1^a=e^{a\ln 1}=e^{0a}$

which sheds a better light on the matter of independence; it is easier to interpret in the familiar terms of "any value" multiplied by zero. Now we know that $0a=0$ for any finite a and that it can be for infinite "values" of a too; notoriously $0\cdot\infty$ is also an indeterminate form, therefore so is $0\frac00$ . So, re-casting the matter as follows:

$1^{\frac00}=e^{0\frac00}$

where it is manifest that the rhs is not independent on the exponent, one may understand that the exponential also isn't wholly independent on the blunt $\frac00$ form and there's no hook or crook in the equality because it relies only on the value of $\ln 1$ and of $e^0$ , no whatsoever singularities. Therefore the rhs can't be equated to 1 as an apodictical certainty and the same goes for the lhs.

Now, without the above manipulations, this was only less obvious than the more common case of multiplication by a factor such as $\frac{x}{x}$ which, Don, is not the same as multiplying by 1; this argument of yours totally forgets the very matter of the indeterminate form. It's the multiplication by an expression which defines a value of 1 only $\forall x\neq 0$ and defines no value for $x=0$ . Even if you remember your own contention, that for $x=0$ the expression "can be any value", you would have to admit that this isn't the same thing as multiplying by 1, or is it?

Don Blazys · 10-23-2008

To: Qfwfq.

When you say that the "indeterminate form":
(0/0) is just as "undefined" as the "undefined operation":
(6/0), then you are dismissing the expression:
(0/0) as "meaningless", when in fact, there are plenty of cases, contexts and circumstances in which equations such as:
(0/0)=1 can be viewed as "meaningfull". (Especially in the context of "limits".)

There are, however, no conditions under which equations such as:
(6/0)= N can be construed as meaningfull, so the expressions:
(0/0), and:
(6/0) must be viewed as "fundamentally different" from each other.

(0/0) "exists" as a logical construct.
(6/0) "does not exist" as a logical construct.

Also, (0/0)= (infinity) if and only if (infinity)*0=0. (Again, the meaning/value of (0/0) depends entirely on the context in which it occured.)

I suppose that different mathematicians have different criteria for what constitutes a "valid result". For me (and most mathematicians my age), those criteria are "consistency" and "beauty".

My garage is the same (consistent) regardless of whether I enter it from the front door, or the back door.

In the same way, at x=1, the equation:

(T/T)a^x=T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1))=

(T/T)a=T(a/T)^((ln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1))=T(a/T),

so that letting T=a results in:

(a/a)a=a(a/a)^(0/0)=a(a/a)=

(1)a=a(1)^(0/0)=a(1),

where, for the sake of consistency, we must conclude that the expressions:

(1) and (1)^(0/0)

are both equal, and "equally trivial" and may therefore be disregarded so as to result in the identity:

a=a=a.

Look at it this way. If, in the above case, we were somehow able to demonstrate that:

(1) and (1)^(0/0)

do not have the exact same value, then we would have to make the further "assumption" that logarithms are inherently "flawed", and must therefore be eliminated from mathematics!

I would then be forever remembered as the "mathematician who demonstrated that the properties of logarithms are bogus".

Like I said before, the proof is merely an unavoidable consequence of "cohesive terms", which are, in turn, merely an extention of the properties of logarithms,... and those properties are definitely not bogus.

Don.

Qfwfq · 10-24-2008

Quote:

Originally Posted by Don Blazys

...there are plenty of cases, contexts and circumstances in which equations such as:
(0/0)=1 can be viewed as "meaningfull". (Especially in the context of "limits".)

Gosh Don, I was talking about the blunt $\frac00$ form and I said so. I was distinguishing it from limits of any kind of expressions in the exponent.

In terms of limits, it is clear that the ratio of two infinitesimals (a distinct thing from the ratio of 0 and 0), can have a limit of any value (according to what the two infinitesimals are). It is also trivial to argue that any limit of:

$1^{f(x)}$

exists and can only be equal to 1. By "any limit of" I mean the limit for x approaching any accumulation point of the domain of $f$ .

$\frac00$ has no domain, no accumulation points, it cannot have any limit. It simply does not and cannot define any value (for any value of $x$ or any other variable).

Do you understand the distinction Don?

Quote:

Originally Posted by Don Blazys

There are, however, no conditions under which equations such as:
(6/0)= N can be construed as meaningfull, so the expressions:

Obviously, if by N you mean a finite value! (or an infinitesimal one) You attempted a sleight of hand here!

Quote:

Originally Posted by Don Blazys

Also, (0/0)= (infinity) if and only if (infinity)*0=0. (Again, the meaning/value of (0/0) depends entirely on the context in which it occured.)

This gets you into the same paralogism I had already pointed out.

Quote:

Originally Posted by Don Blazys

I suppose that different mathematicians have different criteria for what constitutes a "valid result". For me (and most mathematicians my age), those criteria are "consistency" and "beauty".

This is a very dubious statement, self-consistence is a criterion of validity, but beauty is not.

Don Blazys · 10-25-2008

To:Qfwfq.

As mathematicians, we must distinguish between "reasonable implications" and "unreasonable implications".

Now, if N is a non-negative integer, then the "blunt form":

(0/0)=N

implies that any non-negative integer N multiplied by 0 equals 0.

That's a "reasonable implication" because N*0=0 is a "true statement".

However, the "blunt form":

(6/0)=N

implies that any non-negative integer N multiplied by 0 equals 6,

and that's an "unreasonable implication" because N*0=6 is a "false statement".

Thus, the result: (1)^(0/0)=(1)^N=1

is predicated on a "reasonable implication" and a "true statement", so we need not define the exact value of either (0/0) or N.

In fact, if it were somehow concievable that the "blunt form" (0/0)=N defined a "particular value", then that would be a most stupid and ugly thing because it would assign a completely unwarranted significance to that "particular value" and would thereby falsely imply that for the equation:

(1)^(0/0)= (1)^N=1,

some particular value of (0/0) or N is "more true" than the rest!

By contrast, the "result":

(1)^(6/0)=(1)^N

is predicated on an "unreasonable implication" and a "false statement" so we need not even consider it.

Please keep in mind that indeterminate forms such as (0/0) are often initially encountered in their "blunt forms", and that in some cases (such as my proof), they should remain that way, while in other cases, they should be evaluated by introducing more subtle concepts such as limits.

Speaking of limits, it would be easy to introduce the "limit as T approaches c" in my proof, and thereby show that:

z=1 and z=2 results in (1)^(1),

while z>2 results in (1)^(infinity)="indeterminate",

but it would also be quite unnecessary, and would, in my opinion, diminish some of it's generality, simplicity, power, and of course, beauty.

Don.

Qfwfq · 10-27-2008

As a mathematician, I distinguish between "conclusive arguments" and "non sequiturs". Don, I'm also beginning to see that you need to straighten out your grasp on the fundamentals of logic. Let's look at the fallacies in your argument:

Quote:

Originally Posted by Don Blazys

Now, if N is a non-negative integer, then the "blunt form":

(0/0)=N

implies that any non-negative integer N multiplied by 0 equals 0.

That's a "reasonable implication" because N*0=0 is a "true statement".

Actually, it is a truism:

$\frac00=N\Rightarrow N\cdot0=0$

is formally a true implication, but only because the consequent is true and not by force of any argument; you might as well put in any satement as the implicant (even a false one!). There is circularity in attempting to argue the implication; after multiplying both sides of $\frac00=N$ by zero and then applying the implicant to the resulting rhs, you could only conlclude $N\cdot0=N\cdot0$ as a consequent. At this point, in order to deduce $N\cdot0=0$ from that, one would need to use it (i. e. already know it is true).

The trouble is that, since the argument relies on multiplying both sides by zero, even if it could fully lead to the implication ( $\Rightarrow$ ), it certainly couldn't demonstrate a co-implication ( $\Leftrightarrow$ ). It therefore does not justify using the true statement $N\cdot0=0$ to justify $\frac00=N$ at all. I'm under the impression you need to avoid confusion between modus ponens and modus tollens, your argument demonstrates nothing about $\frac00$ defining any value; it is inconlusive.

The implication $\frac{51}{3}=2\Rightarrow N\cdot0=0$ is also formally a true one, what can you deduce from it?

Quote:

Originally Posted by Don Blazys

because N*0=6 is a "false statement".

Of course it is a false satement and I already said so, so we need not even consider it. Friday I left out that a fair comparison would be one between:

$\frac00=N$ and $\frac60=\infty$

where both equalities may be seen as possibly resulting from true limits, by replacement of infinitesimal terms with blunt zeroes. What I meant Friday by "sleight of hand" is that your comparison wasn't a fair one; it's blithering obvious that $\frac60=N$ can't be similarly obtained.

Quote:

Originally Posted by Don Blazys

Please keep in mind that indeterminate forms such as (0/0) are often initially encountered in their "blunt forms", and that in some cases (such as my proof), they should remain that way, while in other cases, they should be evaluated by introducing more subtle concepts such as limits.

I should keep what in mind? You appear not to have understood what I posted on Friday; I was talking about limits, wasn't I? My point was indeed that these things make sense only in the context of computing limits, with the zeroes actually being infinitesimals! Try not to miss my points if you reply to them.

Don Blazys · 10-28-2008

To: Qfwfq

There is no fallacy in my argument.

First of all, let's keep in mind that we are discussing the meanings of the expressions:

(1)^(0/0) and

(1)^(N/0), N>0

as they relate to my proof of the BC.

Thus, if all variables are positive integers and all terms assumed co-prime, then:

(T/T)a^x+(T/T)b^y=(T/T)c^z=

T(c/T)^((zln(c)/(ln(T))-1)/(ln(c)/(ln(T))-1))

and:

((T/T)a^(x/2))^2+((T/T)b^(y/2))^2=((T/T)c^(z/2))^2=

(T(c/T)^(((z/2)ln(c)/(ln(T))-1)/(ln(c)/(ln(T))-1)))^2,

where letting z=1 and z=2 respectively allows us to "cross out" the logarithms, which results in:

(T/T)c=T(c/T), and

((T/T)c)^2=(T(c/T))^2.

Now, my questions to you are:

(1.) Can we now let T=c? (I say yes.)

(2.) Can we "cross out" the logarithms and let T=c if z>2? (I say no.)

(3.) Did letting z=1 and z=2, then "crossing out" the logarithms eliminate the possibility of encountering those pesky "indeterminate forms"? (I say yes.)

(4.) Do those pesky "indeterminate forms" occur if and only if fail to "cross out" the logarithms at z=1 and z=2 before we let T=c? (I say yes.)

(5.) Are those pesky "indeterminate forms" trivial? (I say yes.)

(6.) Do those pesky "indeterminate forms" somehow prevent us from letting T=c at z=1 and z=2? (I say no.)

(7.) Does the possibility of "division by zero" prevent us from letting T=c at z>2? (I say yes.)

(8.) Can we re-define T as "any positive real number other than unity", then show that at T=1 and T=2, the "limit as T approaches c" is c and c^2 respectively? (I say yes.)

Please let me know what your answers to these questions are.

By the way, I disagree that:

(0/0)=N and (6/0)=(infinity)

is a "fair comparison", because while any number:

N*0=0,

there is no number called "infinity" such that:

(infinity)*0=6.

Clearly, if limits are not introduced, then the "blunt" equation:

(0/0)=N

still belongs in the realm of numbers, while:

(6/0)=(infinity)

remains entirely nonsensical.

I also can't see how you can possibly claim that the implication:

51/3=2 (implies) N*0=0

is "formally true". The equation 51/3=2 is false, while the equation N*0=0 is true. How can something false imply, "formally" or otherwise, something true?

Don.

Qfwfq · 10-28-2008

Quote:

Originally Posted by Don Blazys

There is no fallacy in my argument.

First of all, let's keep in mind that we are discussing the meanings of the expressions:

(1)^(0/0) and

(1)^(N/0), N>0

as they relate to my proof of the BC.

This is no way to address an opponents points in debating mathematical or scientific topics.

Quote:

Originally Posted by Don Blazys

Please let me know what your answers to these questions are.

As you have failed to properly address many of my points and reply to my queries, I'm under no obbligation to do so despite that I could. It would not be worth the waste of time and I'm posting during breaks at work and limited time elsewhere and it's time consuming to parse your notation, copy it onto paper more readably and make certain I don't get the parentheses botched up. Folks have other things to attend to, if you fail to be cooperative and more compliant.

Quote:

Originally Posted by Don Blazys

By the way, I disagree that:

(0/0)=N and (6/0)=(infinity)

is a "fair comparison", because while any number:

N*0=0,

there is no number called "infinity" such that:

(infinity)*0=6.

Please note that $0\cdot\infty$ is also notoriously an indeterminate form; the comparison is a fair one. Wasn't that the reason why I said they are equally meaningless?

Quote:

Originally Posted by Don Blazys

Clearly, if limits are not introduced, then the "blunt" equation:

(0/0)=N

still belongs in the realm of numbers, while:

(6/0)=(infinity)

remains entirely nonsensical.

No, they are both meaningless.

Quote:

Originally Posted by Don Blazys

I also can't see how you can possibly claim that the implication:

51/3=2 (implies) N*0=0

is "formally true". The equation 51/3=2 is false, while the equation N*0=0 is true. How can something false imply, "formally" or otherwise, something true?

Don.

Ask a logician. My only remedy to this shall be to suggest you learn the basics of logic; you have fully confirmed your lack of grasp on them; my reply of yesterday to your point:

Quote:

Originally Posted by Don Blazys

That's a "reasonable implication" because N*0=0 is a "true statement".

was overly cautious because it wasn't clear whether or not you meant the very fact that the consequent being identically true is sufficient condition for the implication being true. Apparently this isn't so, you are therefore quite convinced of the argument which I detailed for you in order to expose the fallacies.

Don Blazys · 10-29-2008

To:Qfwfq.

In post#57, you ended your first paragraph with: "Let's look at the fallacies in your argument".

Well, I already know that there are no fallacies in my argument, so the only possible reply from my point of view was and is: "There are no fallacies in my argument."

I suppose that I could also have mentioned that "my grasp of fundamental logic" is quite good, and that from my point of view, it's your grasp of fundamental logic that needs to be "straightened out", but I prefer not to indulge in such childish gibberish.

Neither of us would bother to reply to the other if either of us was truly inept.

Moreover, I never ever considered you to be my "opponent".

I thought that we were exploring my proof together, as friends.

Now, here's where we differ:

I hold that the "blunt" equation:

(1)^(0/0)=1

is both true and correct, while you continue to assert that the expression

(1)^(0/0)

is both "undefined" and "meaningless".

Now, for positive integer variables, and co-prime terms, given the equations:

(T/T)a^x+(T/T)b^y=(T/T)c^z=

T(c/T)^((zln(c)/(ln(T))-1)/(ln(c)/(ln(T))-1)) ___________________(1)

and

((T/T)a^(x/2))^2+((T/T)b^(y/2))^2=((T/T)c^(z/2))^2=

(T(c/T)^(((z/2)ln(c)/(ln(T))-1)/(ln(c)/(ln(T))-1)))^2, ________________(2)

if we let z=1 and z=2 respectively, then the above becomes:

(T/T)a^x+(T/T)b^y=(T/T)c=

T(c/T)^((ln(c)/(ln(T))-1)/(ln(c)/(ln(T))-1)) ________________________(3)

and

((T/T)a^(x/2))^2+((T/T)^b^(y/2))^2=((T/T)c)^2=

(T(c/T)^((ln(c)/(ln(T))-1)/(ln(c)/(ln(T))-1)))^2 _____________________(4)

where the logarithms can now be "crossed out", so that all we have left is:

(T/T)a^x+(T/T)b^y=(T/T)c=T(c/T) ______________________________(5)

and

((T/T)a^(x/2))^2+((T/T)b^(y/2))^2=((T/T)c)^2=(T(c/T))^2 __________(6)

letting T=c in all of the above equations now results in:

(c/c)a^x+(c/c)b^y=(c/c)c=c(1)^(0/0), ___________________________(7)

((c/c)a^(x/2))^2+((c/c)b^(y/2))^2=((c/c)c)^2=(C(1)^(0/0))^2, ______(8)

(c/c)a^x+(c/c)b^y=(c/c)c=c(1) __________________________________(9)

and

((c/c)a^(x/2))^2+((c/c)b^(y/2))^2=((c/c)c)^2=(c(1))^2, ____________(10)

where it is now obvious that the equation:

(1)^(0/0)=1

is perfectly consistent with the cancellation of the logarithms.

If the expression:

(1)^(0/0)

was "undefined", as you claim it is, then both (7) and (8) would become:

a^x+b^y=c= "undefined"

and

a^x+b^y=c^2= "undefined",

and that would make no sense whatsoever, because clearly, defined co-prime equations do indeed exist!

The properties of logarithms can not render all co-prime equations "undefined"!

Thus, it must be the case that:

(1)^(0/0)=1,

and

(1)^(0/0) is not "undefined".

The proof works!

Don.

A Mathematical Emergency.

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